1.
a. `2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`
b.
`n_{Al}=m/M={3,24}/{27}=0,12(mol)`
Theo PTHH ta có: `n_{H_2}=n_{Al}=0,12(mol)`
`⇒ V_{H_2(đktc)}=22,4.n=22,4.0,12=2,688(l)`
c.
Theo PTHH ta có: `n_{Al_2(SO_4)_3}=1/2n_{Al}=1/{2}.0,12=0,06(mol)`
`⇒m_{Al_2(SO_4)_3}=n.M=0,06.342=20,52(g)`
2.
a. `Zn+2HCl->ZnCl_2+H_2`
b. `n_{ZnCl_2}=m/M={20,4}/{136}=0,15(mol)`
Theo PTHH ta có: `n_{ZnCl_2}=n_{H_2}=0,15(mol)`
`⇒V_{H_2(đktc)}=22,4.n=22,4.0,15=3,36(l)`
c. Theo PTHH ta có: `n_{ZnCl_2}=n_{Zn}=0,15(mol)`
`⇒m_{Zn}=n.M=0,15.65=9,75(g)`
`⇒m=9,75(g)`
$\color{red}{\text{#BTS}}$